What is the time complexity of Insert Delete GetRandom O(1) - Duplicates allowed?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Insert Delete GetRandom O(1) - Duplicates allowed?

This problem uses Array, Hash Table, Math, Design, Randomized. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Insert Delete GetRandom O(1) - Duplicates allowed?

Insight 1: Using a list (or array) allows for O(1) random element selection using an index. Insight 2: A hash map can be used to store the values and their corresponding indices in the list, enabling O(1) lookup and removal.

Which companies ask LeetCode 381. Insert Delete GetRandom O(1) - Duplicates allowed in interviews?

This problem is frequently asked at Affirm, Citadel, LinkedIn, Peloton, Yelp and 1 more companies.

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Insert Delete GetRandom O(1) - Duplicates allowed - LeetCode 381 Solution

Insert Delete GetRandom O(1) - Duplicates allowed - Complete Solution Guide

Insert Delete GetRandom O(1) - Duplicates allowed is LeetCode problem 381, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

RandomizedCollection is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element. Implement the RandomizedCollection class: RandomizedCollection() Initializes the empty RandomizedCollection object. bool insert(int val) Inserts an item val into the multiset, even if the item is already present. Returns true if the item is not present, false otherwise. bool remove(i

Detailed Explanation

The problem requires designing a data structure, `RandomizedCollection`, that supports insertion, removal, and random element retrieval in (average) O(1) time, even when duplicate values are present. The `insert` method should return `true` only if the value wasn't present *at all* in the collection, and `false` otherwise. The `remove` method should remove only *one* instance of the value, and return `true` if the value was found and removed, `false` otherwise. `getRandom` should return a random element from the collection, where each element's probability of being chosen is proportional to its frequency in the collection.

Solution Approach

The solution uses a list (or array) `nums` to store the elements of the collection and a hash map `idx_map` to store the mapping between each value and the set of indices where it appears in `nums`. The `insert` operation appends the new element to `nums` and updates `idx_map`. The `remove` operation finds one index of the element to be removed, swaps the element at that index with the last element in `nums`, removes the last element, and updates `idx_map` accordingly. `getRandom` returns a random element from `nums`.

Step-by-Step Algorithm

Step 1: **Initialize**: Create a list `nums` to store the values and a hash map `idx_map` to store the mapping between value and a set of indices.
Step 2: **Insert**: Append `val` to `nums`. Update `idx_map` by adding the index of the newly inserted `val` to the set associated with `val`. Return `true` if `val` was not previously in `idx_map`, otherwise return `false`.
Step 3: **Remove**: If `val` is not in `idx_map`, return `false`. Otherwise, get one index of `val` from `idx_map`. If the index is not the last index, swap `val` at `idx_to_remove` with the last element of nums. Update the idx_map. Finally, remove the last element. If set associated with val is empty, delete the val entry from idx_map.
Step 4: **getRandom**: Generate a random index within the range of `nums` and return the element at that index.

Key Insights

Insight 1: Using a list (or array) allows for O(1) random element selection using an index.
Insight 2: A hash map can be used to store the values and their corresponding indices in the list, enabling O(1) lookup and removal.
Insight 3: Removing an element from the middle of an array takes O(n) time. Swapping the element to be removed with the last element and then popping the last element allows for O(1) removal on average.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Math, Design, Randomized.

Companies

Asked at: Affirm, Citadel, LinkedIn, Peloton, Yelp, Zeta.

Insert Delete GetRandom O(1) - Duplicates allowed | LeetCode 381

Hard

Array Hash Table Math Design Randomized

Time: O(1)Space: O(n)

Solution Code

Available in:

import java.util.*;

class RandomizedCollection {

    private List<Integer> nums;
    private Map<Integer, Set<Integer>> idxMap;

    public RandomizedCollection() {
        nums = new ArrayList<>();
        idxMap = new HashMap<>();
    }
    
    public boolean insert(int val) {
        nums.add(val);
        int idx = nums.size() - 1;
        
        if (!idxMap.containsKey(val)) {
            idxMap.put(val, new HashSet<>());
            idxMap.get(val).add(idx);
            return true;
        } else {
            idxMap.get(val).add(idx);
            return false;
        }
    }
    
    public boolean remove(int val) {
        if (!idxMap.containsKey(val)) {
            return false;
        }

        Set<Integer> valIndices = idxMap.get(val);
        int idxToRemove = valIndices.iterator().next();
        valIndices.remove(idxToRemove);
        int lastIdx = nums.size() - 1;
        
        if (idxToRemove != lastIdx) {
            int lastVal = nums.get(lastIdx);
            nums.set(idxToRemove, lastVal);
            
            idxMap.get(lastVal).add(idxToRemove);
            idxMap.get(lastVal).remove(lastIdx);
        }

        nums.remove(lastIdx);
        
        if (valIndices.isEmpty()) {
            idxMap.remove(val);
        }
            
        return true;
    }
    
    public int getRandom() {
        Random random = new Random();
        return nums.get(random.nextInt(nums.size()));
    }
}

/**
 * Your RandomizedCollection object will be instantiated and called as such:
 * RandomizedCollection obj = new RandomizedCollection();
 * boolean param_1 = obj.insert(val);
 * boolean param_2 = obj.remove(val);
 * int param_3 = obj.getRandom();
 */

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Problem Info

Description

Problem 381: Insert Delete GetRandom O(1) - Duplicates allowed - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Design Randomized

Companies

AffirmCitadelLinkedInPelotonYelpZeta

Stats

Difficulty:Hard

Languages:4

Companies:6

Problem #:381

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: **Initialize**: Create a list `nums` to store the values and a hash map `idx_map` to store the mapping between value and a set of indices.
2
Step 2: **Insert**: Append `val` to `nums`. Update `idx_map` by adding the index of the newly inserted `val` to the set associated with `val`. Return `true` if `val` was not previously in `idx_map`, otherwise return `false`.
3
Step 3: **Remove**: If `val` is not in `idx_map`, return `false`. Otherwise, get one index of `val` from `idx_map`. If the index is not the last index, swap `val` at `idx_to_remove` with the last element of nums. Update the idx_map. Finally, remove the last element. If set associated with val is empty, delete the val entry from idx_map.
4
Step 4: **getRandom**: Generate a random index within the range of `nums` and return the element at that index.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The average time complexity is O(1) because the hash map operations (insert, delete, search) take constant time on average. In the worst-case (e.g., many collisions in the hash table), the hash map operations can take O(n) time, making the remove operation O(n) in the worst case. The space complexity is O(n) because the data structure stores all the elements that have been inserted.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using a Balanced Search Tree

Instead of a hash map, a balanced search tree (e.g., AVL tree, Red-Black tree) could be used to store the elements and their counts. This would provide logarithmic time complexity for insertion and deletion. The array would remain for getRandom.

Time: O(log n) for insert and remove, O(1) for getRandomSpace: O(n)

Trade-offs: This approach would have guaranteed logarithmic time complexity for insert and remove, but would likely be slower in practice due to the overhead of tree operations. The current hashmap approach provides average O(1) complexity, which is generally faster.

✨

Key Insights

1Insight 1: Using a list (or array) allows for O(1) random element selection using an index.
2Insight 2: A hash map can be used to store the values and their corresponding indices in the list, enabling O(1) lookup and removal.
3Insight 3: Removing an element from the middle of an array takes O(n) time. Swapping the element to be removed with the last element and then popping the last element allows for O(1) removal on average.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: Removing an element that doesn't exist. The `remove` method correctly returns `false` if the element is not present in the collection.
Edge case 2: Removing the only instance of an element. The `remove` method correctly removes the element and cleans up the `idx_map` by deleting the key if the set becomes empty.
Edge case 3: Removing the last element in the array. The code handles this case correctly by avoiding unnecessary swaps.
Edge case 4: Inserting duplicate values. The insert method handles the duplicate insertions correctly by keeping track of multiple indices in the idx_map.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate values correctly in the `remove` operation. It's essential to keep track of all indices of each value and remove the element at the correct index.

✗Common mistake 2: Forgetting to update the index map when swapping elements during the `remove` operation, leading to incorrect indexing.

✗Common mistake 3: Not handling edge cases like removing the last element or an element that doesn't exist.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Implementing a weighted random sampler where elements have different probabilities of being selected based on their weights. The RandomizedCollection can be adapted to store the elements and their weights, and getRandom can return a random element based on the weights.

🏦Real-world scenario 2: Load balancing in a distributed system, where different servers have different capacities or loads. The RandomizedCollection can represent the set of available servers, and getRandom can select a server based on its capacity or load.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Insert Delete GetRandom O(1) - Duplicates allowed"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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